Maybe you could see the stereographic projection as a diffeo from the 3-sphere (minus a point) to $\mathbb{R}^3$, and then transport the Hopf fibration to $\mathbb{R}^3$. That still leaves one manifold homeomorphic to $\mathbb{R}$ and not to $\mathbb{S}_1$, because of the point we eliminated at the beginning.
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D. ThomineMar 3 '12 at 14:02

I've found a way to represent $\mathbb{R}^3$ as a disjoint union of manifolds diffeomorphic to $\mathbb{S}_1$, but it is not very elegant and it makes no use of the stereographic projection. Are you interested?
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D. ThomineMar 7 '12 at 15:30

1 Answer
1

Since you said you're interested in D. Thomine's answer (and s/he seems to have forgotten about it), here's an answer using no complex analysis or stereographic projection. We will use transfinite induction. We will show $\mathbb{R}^3$ can be partitioned into circles of radius $1$. This idea actually given as an exercise in Ciesielski's book Set Theory for the Working Mathematician. (In the same book, he proves that $\mathbb{R}^2$ can't be partitioned as a union of circles of positive radius.)

To that end, choose a bijection with $\mathbb{R}^3$ and $\mathfrak{c}$, the cardinality of the continuum. (Equivalently, well order $\mathbb{R}^3$ with its minimal well ordering.) The important thing about $\mathfrak{c}$ is that it is a cardinal number, meaning for any ordinal number $\alpha < \mathfrak{c}$, we have $|\alpha| < |\mathfrak{c}| = \mathfrak{c}$ where $|\cdot |$ denotes cardinality.

To begin the induction, let $r_0$ denote the "first" real number. Choose any circle $C_0$ of unit radius which goes through $r_0$.

Now, assume inductively that for all $\beta < \alpha$, we have chosen pairwise disjoint circles of radius $1$ so that all of the $r_\beta$ lie on 1. We wish to extend the induction to $r_\alpha$.

First, if $r_\alpha$ already lies on a prechosen circle, we are done. So, we may assume $r_\alpha$ does not lie on any previously chosen circle of radius 1.

Now, consider all the planes in $\mathbb{R}^3$ passing through $r_\alpha$. It is easy to see that there are $\mathfrak{c}$ such planes. Since, at this point, we have chosen at most $|\alpha|$ circles and $|\alpha|< \mathfrak{c}$, and each circle lies in one plane, there must be a plane $P_\alpha$ which doesn't contain any of our previously chosen circles (though, of course, they may intersect it in $1$ or $2$ points).

Let's just focus on $P_\alpha$ for now. Consider all the circles of radius $1$ contained in $P_\alpha$ passing through $r_\alpha$, which I'll call candidate circles. Again, a not-too-hard counting argument shows there are $\mathfrak{c}$ candidate circles. Each of our previously chosen circles intersects $P_\alpha$ in at most 2 points so there are at most $2|\alpha| = |\alpha| < \mathfrak{c}$ "bad" points in $P_\alpha$ which we must avoid. Any point in $P_\alpha$ is on at most 2 candidate circles, so all "bad" points remove at most $2|\alpha| < \mathfrak{c}$ candidate circles from consideration. But since there are $\mathfrak{c}$ candidate circles, there must be at least one left over. Let $C_\alpha$ be one of these left overs candidates.

By "candidateness", $r_\alpha \in C_\alpha$. Further, $C_\alpha$ cannot intersect any of our previously chosen circles because those intersection points would correspond to "bad" points, which we avoided. Thus, we have continued the induction.